Answer:
2.08335 kj
Explanation:
Given data:
Mass of lead = 85 g
Initial temperature = 200°C
Final temperature = 10°C
Heat lost = ?
Cp of lead = 0.129 j/g.°C
Solution:
Specific heat capacity: Cp
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 10°C - 200°C
ΔT = -190°C
Now we will put the values:
Q = 85 g × 0.129 j/g.°C × -190°C
Q = 2083.35 j
Joule to kilo joule:
2083.35/1000 = 2.08335 kj
